Lösung 3.4:2a

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The left-hand side is "
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The left-hand side is "2 raised to something", and therefore a positive number regardless of whatever value the exponent has. We can therefore take the log of both sides,
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<math>\text{2}</math>
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raised to something", and therefore a positive number regardless of whatever value the exponent has. We can therefore take the log of both sides,
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{{Displayed math||<math>\ln 2^{x^2-2} = \ln 1\,,</math>}}
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<math>\ln 2^{x^{2}-2}=\ln 1</math>
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and use the log law <math>\ln a^b = b\cdot \ln a</math> to get the exponent <math>x^2-2</math> as a factor on the left-hand side,
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and use the log law
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{{Displayed math||<math>\bigl(x^2-2\bigr)\ln 2 = \ln 1\,\textrm{.}</math>}}
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<math>\lg a^{b}=b\centerdot \lg a</math>
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to get the exponent
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<math>x^{\text{2}}-\text{2 }</math>
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as a factor on the left-hand side
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Because <math>e^{0}=1</math>, so <math>\ln 1 = 0</math>, giving
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<math>\left( x^{\text{2}}-\text{2 } \right)\ln 2=\ln 1</math>
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{{Displayed math||<math>(x^2-2)\ln 2=0\,\textrm{.}</math>}}
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This means that ''x'' must satisfy the second-degree equation
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Because
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{{Displayed math||<math>x^2-2 = 0\,\textrm{.}</math>}}
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<math>e^{0}=1</math>, so
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<math>\text{ln 1}=0</math>, giving:
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Taking the root gives <math>x=-\sqrt{2}</math> or <math>x=\sqrt{2}\,</math>.
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<math>\left( x^{\text{2}}-\text{2 } \right)\ln 2=0</math>
 
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Note: The exercise is taken from a Finnish upper-secondary final examination from March 2007.
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This means that
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<math>x</math>
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must satisfy the second-degree equation
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<math>\left( x^{\text{2}}-\text{2 } \right)=0</math>
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Taking the root gives
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<math>x=-\sqrt{2}</math>
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or
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<math>x=\sqrt{2}.</math>
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NOTE: the exercise is taken from a Finnish upper-secondary final examination from March 2007.
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Version vom 11:15, 2. Okt. 2008

The left-hand side is "2 raised to something", and therefore a positive number regardless of whatever value the exponent has. We can therefore take the log of both sides,

Vorlage:Displayed math

and use the log law \displaystyle \ln a^b = b\cdot \ln a to get the exponent \displaystyle x^2-2 as a factor on the left-hand side,

Vorlage:Displayed math

Because \displaystyle e^{0}=1, so \displaystyle \ln 1 = 0, giving

Vorlage:Displayed math

This means that x must satisfy the second-degree equation

Vorlage:Displayed math

Taking the root gives \displaystyle x=-\sqrt{2} or \displaystyle x=\sqrt{2}\,.


Note: The exercise is taken from a Finnish upper-secondary final examination from March 2007.

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_3.4:2a“